I've been thinking about Grassmannians recently. Think of $\mathbb{R}^k$ as a $k$-dimensional vector space. Let $\text{Gr}_n(\mathbb{R}^k)$ denote the Grassmannian of all $n$-dimensional vector subspaces of $\mathbb{R}^k.$ (This is a compact, Hausdorf topological manifold of dimension $n(k-n)$.) Let

There's a standard idea of a vector bundle $\pi : \Gamma^n(\mathbb{R}^k) \twoheadrightarrow \text{Gr}_n(\mathbb{R}^k)$ given by $\pi(X,v) := X.$ This bundle has the nice property that lots of other bundles can be realised as sub-bundles of it. There is a more general definition, where we use $\mathbb{R}^{\infty}$ in place of $\mathbb{R}^k$. My question is about why we define $\mathbb{R}^{\infty}$ the way we do.

We define $\mathbb{R}^{\infty}$ as the set of infinite sequences $(x_1,x_2,x_3,\ldots)$ where each $x_i \in \mathbb{R}$ and only finitely many of the $x_i$ are non-zero. We identify $\mathbb{R}^k$ with the sequences of the form $(x_1,\ldots,x_k,0,0,\ldots),$ and then topologize $\mathbb{R}^{\infty}$ as the direct limit of the sequence $ \mathbb{R}^1 \subset \mathbb{R}^2 \subset \mathbb{R}^3 \subset \ldots$ Then we get the universal bundle $\pi : \Gamma^n(\mathbb{R}^{\infty}) \twoheadrightarrow \text{Gr}_n(\mathbb{R}^{\infty}).$

My question is why do we insist that only finitely many of the $x_i$ are non-zero for each $(x_1,x_2,x_3,\ldots) \in \mathbb{R}^{\infty}$? I understand that it gives a countably infinite dimensional vector space, but that's a result of the definition; it doesn't explain why we define it the way we do. I suspect that it's related to the topology, but I don't really know.

Also it may be good to keep in mind, that this setup gives $$Gr_n(\mathbb{R}^\infty)= \bigcup Gr_n(\mathbb{R}^k)$$ as a directed union. So it is well approximated by finite dim objects.
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Donu ArapuraAug 19 '11 at 22:25

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@Ryan I think the question is interesting for the beginning graduate student. It raises a lot interesting questions about topology and limits. Even though most of us have thought through these, I see no harm in leaving it open, unless the OP wants to accept the answers in the comments.
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Scott CarterAug 19 '11 at 22:33

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@Scott: I agree some beginning grad students might be interested in this question, but that's far from making it an appropriate MO question.
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Ryan BudneyAug 19 '11 at 22:55

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"Why not spend more time trying to understand what people are asking?" I could turn this round and say "why not spend more time crafting a more localized question that is less open-ended?" Note that I did not vote to close your question. The fact that the answer you wanted is "It's a CW complex" is only obvious with hindsight. People work with spaces that are not CW complexes, so how is one meant to have known that this was the desired answer? I reiterate that while your original question may have been clear to you, it was not clear what kind of answer you were expecting or desiring.
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Yemon ChoiAug 21 '11 at 4:33

5 Answers
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This way it's a CW-complex, so studiable by standard tools of algebraic topology (pretty close to Donu's comment).

For a striking example of how these things can depend on the
definition, look at the infinite unitary groups $U(\infty)$
vs. $U({\mathbb H})$, where the first is defined as $\bigcup U(n)$
and the second as the group of all unitary operators on Hilbert space.
Then $U(\infty)$ has very interesting homotopy
($\pi_n(U(\infty)) = 0$ if $2|n$, $= {\mathbb Z}$ if not),
whereas $U({\mathbb H})$ is contractible.

I must admit that I'm not happy with the example you chose! $U(\infty)$ is not the unitary group of $\sum_{\infty} \mathbb{R}$. You can take a Hilbert space $\mathbb{H}$ and form a very nice closed subgroup of $U(\mathbb{H})$ which has the homotopy type of $U(\infty)$. What's more, this one is a smooth (infinite dimensional) manifold, unlike $U(\infty)$. There are lots of models for $BU(\infty)$, each with their own advantage, but I think that several of them are CW complexes. The advantage of this model that springs to mind is that it is fairly easy to show that (ctd)
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Loop SpaceAug 20 '11 at 7:39

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Thank you very much Allen, that makes sense. And thanks to Donu too; I wasn't saying he was wrong, I was just expressing a confusion. I appreciate both of your efforts. I'm glad I managed to get an answer despite other people's efforts to shut me out.
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Fly by NightAug 20 '11 at 18:29

@Akhil: That is the content of Kuiper's theorem.
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Ulrich PennigAug 22 '11 at 16:20

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Akhil: Kuiper's paper is well worth a read on that. One key step is that you can write a Hilbert space as a countable sum of copies of itself. Then if you consider a map in to the unitary group from a sphere, you can shift it so that it takes values in just one factor, then do the Mazur-swindle (is that it's name? I forget) which cancels what happens in that factor by something in the next, then that gets cancelled by something in the third, and so on. It really is a neat proof.
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Loop SpaceAug 22 '11 at 18:03

There's been quite a bit of discussion of this thread in the meta thread. I'd like to take a stab at answering my interpretation of what the question is getting at. In a sense there's at least two things going on, and that's part of why I think there's been so much discussion.

To take a step back from Grassmannians, I'd like to mention why one might be interested in $\mathbb R^\infty$. The Whitney embedding theorem states that every continuous function $f : N \to \mathbb R^k$ can be approximated uniformly by a $C^\infty$-smooth embedding provided $k \geq 2n+1$ where the dimension of $N$ is $n$. Moreover, any two embeddings $N \to \mathbb R^k$ are isotopic provided $k \geq 2n + 2$. So if you wanted to, you could replace the class "$n$-manifolds up to diffeomorphism" with "isotopy classes of $n$-dimensional submanifolds of $\mathbb R^k$ provided $k \geq 2n+2$".

A key nice result about the weak topology on $\mathbb R^\infty$ is that any continuous function from a compact space to $\mathbb R^\infty$ has an image in $\mathbb R^k$ for some $k$. So from the perspective of the Whitney embedding theorem above, "$n$-dimensional manifolds up to diffeomorphism" is precisely "$n$-dimensional submanifolds of $\mathbb R^\infty$ up to isotopy". The key thing here is the ambient space is now independent of the dimension of the manifold you're talking about. This is pretty much exactly what's going on with the Grassmannians.

Given a vector bundle $p : E \to B$ over a finite-dimensional space $B$ (say a manifold or a CW-complex), there exists a classifying map for the bundle, meaning $p$ is isomorphic to the pull-back of the tautological bundle over $G_{n,k} \equiv G_k(\mathbb R^n)$. $n$ is just some sufficiently large integer. Although people don't state it this way, this theorem is basically the Whitney embedding theorem but for vector bundles. Because it's saying that up to isomorphism, $E$ is a collection of pairs $(b,v)$ where $b \in B$ and $v \in \chi^{-1}(h(b))$, where $\chi : E_{n,k} \to G_{n,k}$ is the tautological bundle over $G_{n,k}$. $h : B \to G_{n,k}$ the classifying map. In a sense we've "embedded" the vector bundle in Euclidean space, well, we've made the fibers as subspaces of Euclidean space.

But again, you have the "for some $n$ sufficiently large" thing. And like with manifolds $n$ has an upper bound in terms of the dimension of $B$ (if $B$ is finite dimensional). Since it's sometimes awkward to carry-around these "for sufficiently large $n$" statements, you take the limit space $G_{\infty,k}$ and now your statement is far more clean, because any $k$-dimensional vector bundle over any space $B$ is the pull back of some map $B \to G_{\infty,k}$. The point is that $G_{\infty,k}$ is a universal space -- independent of $B$ or the vector bundle over $B$. The weak topology is exactly what allows us to ensure this happens.

Perhaps the following is so obvious that no one saw fit to mention it in a comment:

Let $X$ be a paracompact topological space. With the definition the way it is, homotopy classes of maps from $X$ to the $n$-Grassmannian are in bijection with isoclasses of rank $n$ vector bundles on $X$. (If you haven't got to them yet, this is thms 5.6 and 5.7). This is what really gets used in application to Stiefel-Whitney classes.

I don't know what is classified by the other definition of the $n$-Grassmannian you suggest ($n$ dimensional subspaces of the product of infinitely many copies of the reals). But certainly for your purposes (working through Milnor+Stasheff) this is the point.

The fact that the usual Grassmannian is a direct limit is used in several technical lemmas in chapter 5, but I have never tried pushing the construction you suggest through.

That result only depends on the homotopy type of the Grassmannian, and all the different possibilities have the same homotopy type, so it doesn't address the actual question as to why Milnor and Stasheff use this particular model.
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Loop SpaceAug 22 '11 at 18:04

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Andrew, I thought the question was pretty clear (and don't understand all the fuss): why define the Grassmannian of n-planes using the direct sum instead of direct product? The cleanest possible answer is: because using direct product doesn't classify vector bundles on paracompact spaces. Unfortunately, it's not clear to me that this is true, but it might be worth thinking about for a little while for someone who is interested.
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Stephen GriffethAug 24 '11 at 12:55

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SPG I didn't (and still don't) see anything that implies that we are comparing the direct sum with the direct product. It seems more "Why the direct sum as opposed to ... err ... anything else?". There certainly are other things that work just as well. (I don't know, either, whether the direct product is one of them.) But that's one of the issues I have with this question: it is possible to read it in too many different ways and each has a subtly different answer.
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Loop SpaceAug 26 '11 at 13:32

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From the question: "My question is why do we insist that only finitely many of the xi are non-zero for each (x1,x2,x3,…)∈R∞?" If we do not insist this, we get the direct product, do we not? This seems unambiguous to me.
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Stephen GriffethAug 29 '11 at 14:40

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In fact, that's the only sentence in the OP's question with a question mark after it, so I'm even more puzzled by your assertion that you don't see anything that implies we are comparing the direct sum with the direct product.
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Stephen GriffethAug 29 '11 at 14:43

Obviously tastes/opinions vary, but I think some ambiguous, or insufficiently localized, or not publishable-research-y enough, but nevertheless valuable to (a significant demographic of) research mathematicians... In fact, sometimes these questions are exactly the "dumb, non-research" questions that "everyone" (anyway, many people) have asked themselves... and not received a cogent answer.

One cliched-but-important (in my opinion) point is the "naive category-theory" explanation of why $\mathbb R^\infty$ is "defined" to be what it is. This does raise the entirely legitimate meta-meta-question of why we "have definitions", and "who is authorized to make them"... to which the easy answer (in my opinion, with some hindsight) is that, not merely must definitions capture the phenomena of important examples (or else the definitions are dumb), but, actually, as it seems to happen very often, re-ordering the "definition...theorem" sequence to "(mapping-)characterization..." rewrites the narrative so that the required/desired property is written in mildly category-theoretic terms, and the technical bit is perhaps proof-of-existence.

That is, a (typically, set-theoretic) "definition" is actually just _one_specific_construction_ of an object whose important features (if it exists at all) are completely determined by its interactions with other objects. That is, its characterization is "category theoretic" rather than "set theoretic".

(Yes, this is an advertisement for a certain little bit of category theory, though it is not a paid advertisement, insofar as I do not at all advocate formal category theory, nor would I advocate allocating one's personal resources to fretting over reconciliation of set theory and category theory... e.g., Grothendieck's "universes" and large cardinals? Fun, but likely not refering directly to one's original issue...)

So, rewriting the question about "why is the definition of $\mathbb R^\infty$ what it is?", we are required to ask what function this thing should have. Well, it is almost immediate that it should be the _ascending_union_ of the $\mathbb R^n$'s. That is, (upon reflection!) it is a (filtered) colimit (a.k.a. "inductive limit"). That is, it should/must/does have certain diagrammatic/mapping properties... as opposed to goofy set-theoretic constructional details.

Issues about infinite-dimensional Grassmannians... infinite-dimensional simplicial complexes... do share that basic feature, namely, that there is a mapping property (if only ascending-union sorts of (filtered) colimit properties) that are relevant.

Truly, the above viewpoint seems to me to be extraordinarily efficient/effective as explanatory device...

(And, one more time, questions that fail to be "documentable research" sometimes are far more interesting and useful to "us" than more focused ones... Of course, this is not a general rule...)

If we don't insist that all but finitely many entries are non-zero, then we get the (set-theoretic) product of countably many copies of ${\mathbb R}$. If we look at the degree-n Grassmanian of this object, then this is not the object defined in Milnor Stasheff. Hence there exists at least one theorem that is true for the object they define, which is not true for the object we have just defined.

I very rarely downvote. However, this answer is a role model for an unuseful answer. (Which may make it of some use, after all.)
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Gil KalaiAug 30 '11 at 9:00

Yemon, your answer may be intended rhetorically but is it evidently correct? Could it be that the Grassmannians of $n$-dimensional subspaces of the sum and the product, with their tautological vector bundle, are isomorphic in the category of spaces with vector bundles?
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Tim PerutzAug 30 '11 at 16:30